Answer:
C would be one space to the right of D
B would be two spaces below D
A would be three spaces to the left from the new location B
Step-by-step explanation:
If you take the derivative of your equation, you get:
2′″−″−′+′=0
2
y
′
y
″
−
x
y
″
−
y
′
+
y
′
=
0
or
″(2′−)=0.
y
″
(
2
y
′
−
x
)
=
0.
Let =′
v
=
y
′
and we have ′(2−)=0,
v
′
(
2
v
−
x
)
=
0
,
so either ′=0
v
′
=
0
and =
v
=
c
or =/2
v
=
x
/
2
.
Then ′=
y
′
=
c
and so =+
y
=
c
x
+
d
or ′=/2
y
′
=
x
/
2
and =2/4.
y
=
x
2
/
4.
Plugging the first into the original equation gives =−2
d
=
−
c
2
. So there are two solutions =−2
y
=
c
x
−
c
2
for some constant
c
and =2/4
y
=
x
2
/
4
. I don't know if this is all the solutions
For resting your heart would beat 72 times per minute and for running your heart would beat 150 times per minute (and together it would be 222)
30(30 – x) = 540
distribute
900 -30x = 540
subtract 900 from each side
-30x = -360
divide by -30
x =12
Step-by-step explanation:
The FORTRAN program "cascade" (see Appendix) can be used to evaluate the performance of TSC plans cascaded through time.
